Question

In an acute angled triangle ABC, the minimum value of tan^n A + tan^n B + tan^n C. is
(when ne N, n > 1)
​

Answer 1

$\large\underline{\sf{Solution-}}$

We know,

In acute triangle ABC,

$\rm :\longmapsto\:A + B + C = \pi$

can be rewritten as

$\rm :\longmapsto\:A + B = \pi - C$

Apply tan on both sides, we get

$\rm :\longmapsto\:tan(A + B) =tan( \pi - C)$

We know,

$\boxed{ \bf{ \: tan(x + y) = \frac{tanx + tany}{1 - tanxtany}}}$

So, using this we get

$\rm :\longmapsto\:\dfrac{tanA + tanB}{1 - tanAtanB} = - tanC$

$\rm :\longmapsto\:tanA + tanB = - tanC + tanAtanBtanC$

$\rm :\implies\:tanA + tanB + tanC = tanAtanBtanC$

Now, we kmow that

Arithmetic mean is always greater than or equals to geometric mean.

So, apply this property on tanA, tanB and tanC, we get

$\rm :\longmapsto\:\dfrac{tanA + tanB + tanC}{3} \geqslant \sqrt[3]{tanAtanBtanC}$

$\rm :\longmapsto\:tanA + tanB + tanC \geqslant 3 \sqrt[3]{tanAtanBtanC}$

On cubing both sides, we get

$\rm :\longmapsto\: {(tanA + tanB + tanC)}^{3} \geqslant 27tanAtanBtanC$

As we proved above,

$\red{\rm :\longmapsto\:\:tanA + tanB + tanC = tanAtanBtanC}$

So, using this, we get

$\rm :\longmapsto\: {(tanAtanB tanC)}^{3} \geqslant 27tanAtanBtanC$

$\rm :\longmapsto\: {(tanAtanB tanC)}^{2} \geqslant 27$

$\red{\rm :\implies\:tanAtanBtanC \geqslant 3 \sqrt{3} }$

or

$\red{\rm :\implies\:tanA + tanB + tanC \geqslant 3 \sqrt{3} }$

Now, we have to evaluate minimum value of

$\rm :\longmapsto\: {tan}^{n}A + {tan}^{n}B + {tan}^{n}C$

Now, we know from property of power of Arithmetic mean that

$\boxed{ \bf{ \: {x}^{n} + {y}^{n} \geqslant {(x + y)}^{n}}}$

So, using this property, we get

$\rm :\longmapsto\: {tan}^{n}A + {tan}^{n}B + {tan}^{n}C \geqslant {(tanA + tanB + tanC)}^{n}$

As we proved above,

$\red{\rm :\longmapsto\:\:tanA + tanB + tanC \geqslant 3 \sqrt{3} }$

So, using this, we get

$\rm :\longmapsto\: {tan}^{n}A + {tan}^{n}B + {tan}^{n}C \geqslant {(3 \sqrt{3} )}^{n}$

Hence,

Minimum value of

$\rm :\longmapsto\: {tan}^{n}A + {tan}^{n}B + {tan}^{n}C = {(3 \sqrt{3} )}^{n}$